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INTERNATIONAL ISLAMIC UNIVERSITY
MALAYSIASEM I 2005
ECE 1101
EXPERIMENT NO 3
TITLE OF EXPERIMENT:
THEVENIN’S & NORTON’S THEOREMS
POWER TRANSFER & SUPERPOSITION
THEOREMS
16 AUGUST 2005
KHAIRUL EINUDDIN ABD RAHIM (0438945)MOHD HAKIKI MD TOHID (0434459)
SOIFFOINE ISSA SAID (0519749)LAB SECTION: 3
LAB SESSION: (TUESDAY, 4.30PM – 7.30PM)LECTURER : SISTER FARAH AYU
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THEVENIN’S & NORTON’S THEOREMS
6:59 With Him are the keys of the unseen, the treasures that none knoweth but He. He
knoweth whatever there is on the earth and in the sea. Not a leaf doth fall but with Hisknowledge: there is not a grain in the darkness (or depths) of the earth, nor anythingfresh or dry (green or withered), but is (inscribed) in a recorded clear (to those who canread)
INTRODUCTION
This experiment basically is about the Norton’s and Thevenin’s Theorems:
- Thevenin’s theorem states that a linear two-terminal circuit can be replaced by anequivalent circuit consisting of a voltage source V Thevenin’s in series with a
resistor R Thvenin’s, where V Thevenin’s is the open-circuit voltage at theterminals and R Thevenin’s is the input or equivalent resistance at the terminalswhen the independent sources are turned off.
- Norton’s theorem states that a linear two-terminal circuit can be replaced by anequivalent circuit consisting of a current source In in parallel with a resistor R Norton’s , where I Norton’s is the short-circuit current through the terminal andR Norton’s is the input or equivalent resistance at the terminals when theindependent sources are turned off.
APPARATUS
- DC supply (Vs=15V)- Digital multimeter.- Resistors; R1=1.8kΩ , R2=3.6kΩ, R3=820Ω, R4=R5=100Ω, R L=180Ω.- Pspice software
OBJECTIVE
1. To find the current flowing in a particular resistor (variable load) of a network byapplication of Thevenin’s and Norton’s theorems.
2. To verify the theorems by comparing the simulated values to those obtained by
measurement. PROCEDURE
(a) Thevenin’s Theorem:
1. R L is selected as the resistor where it is propose to determine the current value.
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2. The supply voltage and resistance of each resistor has been measured. Thesevalues are recorded in Table 4-1.
3. Contact up the circuit as shown in Figure 4-1. Do not turn on the supply.4. Resistor RL is removed from the network.5. The supply has been turned on. The voltage between the points A and D of the
network has been measured. This is the Thevenin’s voltage. The voltage valuehas been recorded.6. The supply has been switched off. The supply has been replaced with a short
circuit.7. The resistance between the terminals A and D has been measured. This is the
Thevenin’s resistance. The value has been recorded.8. The resistor R5 is being placed back in circuit with an ammeter between
terminals A and B or C and D.9. The supply is being placed back in the circuit instead of the short circuit.10. The supply has been turned on back. The current value flowing in the resistor
R5 has been read and recorded.
11. Using the values of the supply voltage and the resistors as measured, draw twoPSpice schematics has been drawn and simulated to find the theoretical valuesof thevenin’s resistance, thevenin’s voltage and current through RL.
(b) Norton’s Theorem:
1. R is selected as the resistor where it is proposed to determine the current value.2. Contact up the circuit as shown in Figure 4-1. Do not turn on the supply3. Resistor RL has been removed from the network.4. The supply has been turned on. The current shown by the ammeter between
terminals a and d has been read. This is Norton’s current, In. Its value has beenrecorded.5. The supply has been switched off. The supply has been replaced with a short
circuit.6. The resistance between the terminals a and d has been measured. This is
Norton’s resistance. The value has been recorded.7. The resistance R L has been placed back in circuit with an ammeter between
terminals a and b or c and d.8. Instead of the short circuit, the supply in circuit has been placed back into the
circuit.9. The switch S has been closed. The current value flowing in the resistor R L has
been read and recorded.10. Using the values of the supply voltage and the resistors as measured, a PSpiceschematic has been drawn and simulated to find the theoretical values of shortcircuit Norton’s current. Norton’s resistance is equal to Thevenin’s resistance,and the same goes for current through RL.
11. A Norton’s equivalent circuit inclusive of resistor RL has been drawn.
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CIRCUIT DIAGRAMS
(a)Thevenin’s theorem
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(b) norton’s theorem
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DISCUSSION
1. This experiment had been conducted to find the current flowing in a particular resistor (variable load) of a network by application of Thevenin’s and Nortontheorems.
2. Thevenin’s theorem states that a linear two-terminal circuit can be replaced by anequivalent circuit consisting of a voltage source V Thevenin’s in series with aresistor R Thvenin’s, where V Thevenin’s is the open-circuit voltage at theterminals and R Thevenin’s is the input or equivalent resistance at the terminalswhen the independent sources are turned off.
3. Thevenin’ voltage is determined by measure the voltage between A and D whenthe resistor RL is remove from the circuit.
4. The Thevenin’s resistance can be measured by replace the supply with a shortcircuit, and then measured the value of resistance between A and D.5. Norton’s theorem states that a linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a current source In in parallel with a resistor R Norton’s , where I Norton’s is the short-circuit current through the terminal andR Norton’s is the input or equivalent resistance at the terminals when theindependent sources are turned off.
6. Norton’s current is determine by remove the resistor RL from the network andturn on the supply and then read the current through terminal a and d.
7. To measure the Norton’s resistance, the supply must be replace with a shortcircuit and measure the resistance between the terminal a and d.
8. During conduct this experiment, there are some errors which have effect thevalue of the variable. The errors are:• The value of the current voltage is not correct. This happen because the
ammeter is damaged.• The measured value is not the actual value. During measured the value, the
value is not stable. So, we just take value which is likely to be accurate.9. To minimize the error,
• We must assure that the ammeter is working by check it before use it.• Take many readings then take the average value of it to minimize the error.
10. In general, the initial objective of our experiment is achieved.
ANALYSIS, DEDUCTIONS AND CONCLUCIONS
1. The aims of this experiment have been achieved because the Thevenin’s and Norton theorem have been verified.
2. The value from the PSpice is differentiating with the value from this experiment.The small percentage errors indicate that the theorems have been verified.
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3. Some errors occur while handling this experiment which leads to inaccurateresult compared with experiment using Pspice. But the error is small and the percentage is low. The error occur because of ;• The value of the supply voltage is not consistence. This happen because
when we want to measured the other variable value, the supply voltage will
decrease• The value of the resistor is not same as the printed one. May be the resistor
has been used for long time, so the value had been decreased• The measured value is not the actual value. During measured the value, the
value is not stable. So, we just take value which is likely to be accurate.4. Kirchoff’s laws are not applicable for some transistor circuits.5. Transistor circuit.
CALCULATIONS
All theoretical values were generated using PSpice software.
Percentage error:
% error = |Theoretical value-experimental value|---------------------------------------------- x 100%
Theoretical value
(a)Thevenin’s Theorem:Percentage difference:
i. Thevenin’s resistance: 618.2 Ω - 614Ω × 100 =0.68 %
618.2 Ω
ii. Thevenin’s voltage: 4.88 V – 4.81V × 100 =1.43%4.88 V
iii. Current in R L: 5.623 mA –6.01 mA × 100 = 6.88%5.623 mA
(b) Norton’s Theorem:Percentage difference:
i. Norton’s resistance: : 618.2 Ω -614 Ω × 100 =0.68 %618.2Ω
ii. Norton’s current: 6.401 mA –7.63 mA × 100 =19.2 %6.401 mA
iii. Current in R L: 5.623 mA –6.01 mA × 100 = 6.88%
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5.623 mA
REFERENCE
1. Introductory Circuit Analysis, Robert L. Boylestad.Prentice-Hall International, Inc. 8th Edition, 1994.
2. Fundamentals of Electric Circuit, Charles K. Alexander, Matthew N.O. Sadiku.McGraw-Hill Higher Education. 2nd Edition, 2003.
POWER TRANSFER & SUPERPOSITION
THEOREMS
Test (a) maximum power transfer:
AIMS
For the theory part, maximum power is transferred to the load when the loadresistance equals the Thevenin resistance as seen from the load. To this experiment,we are going to verify the maximum power transfer theorem.
APPARATUS
1- DC power supply = 15V2- Digital millimeters.3- Variable load resistor R L
Using decade resistance box Fixed resistor R th = 5.1kΩ.
4- PSpice software.
METHOD
In this experiment we are going to develop several tests to verify the maximum power transfer theorem. Firstly, we construct the Thevenin equivalent circuit in figure5.1. After turn off the supply voltage, we set the slider of resistor RL for maximumresistance. Then, turn on the supply voltage. In line to this, we take the correspondingreading of voltage and current for various values of load resistance RL (1kΩ each step)including the condition when RL is short circuited (RL=0). For the theoretical value, weused the PSpice software.
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CIRCUIT DIAGRAMS
RESULT
Sourcevoltage(volts)
Sourcepoweroutput(watt)
Loadvoltage(volt)Theo
Current(A)theo
Loadpower(watt)
exp
Loadresistance
(kΩ)
Transferefficiency(%) exp
15 41.1 0.00 2.941 0.00 0
15 33.9 2.459 2.459 6.04 1
15 31.05 4.225 2.113 8.99 2
15 26.7 5.556 1.852 10.28 3
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15 24 6.593 1.648 10.86 4
15 21.45 7.426 1.485 11.03 5
15 19.65 8.108 1.351 10.96 6
15 17.85 8.678 1.240 10.76 7
15 16.65 9.160 1.145 10.49 8
15 15.45 9.57 1.064 10.19 9
15 14.4 9.93 0.993 9.87 10
Transfer Efficiency
Transfer efficiency for R L 0 kΩ = 0.00 x 100 = %45.00
Transfer efficiency for R L 1 kΩ = 6.04 x 100 = %33.9
Transfer efficiency for R L 2 kΩ = 8.99 x 100 =%31.05
Transfer efficiency for R L 3 kΩ = 10.28 x 100 =%26.7
Transfer efficiency for R L 4 kΩ = 10.86 x 100 =%24
Transfer efficiency for R L 5kΩ = 11.03 x 100 =%21.45
Transfer efficiency for R L 6 kΩ = 10.96 x 100 =%19.65
Transfer efficiency for R L 7kΩ = 10.76 x 100 =%
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17.85
Transfer efficiency for R L 8 kΩ = 10.49 x 100 =%16.65
Transfer efficiency for R L 9 kΩ = 10.19 x 100 =%15.45
Transfer efficiency for R L 10 kΩ = 9.87 x 100 = %14.4
Test (b) – The superposition theorem
AIMS
As the theory of the experiment, the superposition principles states that thevoltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (Or currents through (the element due to each independent sourceacting alone. In this experiment, we are going to develop several test that can verify thesuper position theorem as stated by the theorem.
APPARATUS
1-Two DC supplies (6V and 12V)2-Digital multimeter.
3-Resistor R 1 = 470Ω R 2 = 5.6kΩ R 3 = 3.6kΩ
4--PSpice software.
METHOD
In this experiment, we have developed several tests to achieve our aim. Firstly,we measure the resistance of each resistor and the voltage of each source and fill thetable 5.2. Then, switches S1 in position A and S2 in position D. we then record the valuein the table 5-3. Don’t forget to refer to the last configuration in figure 5-4. In line tothis, a voltmeter is placed across R2 accordingly and the node voltage is recorded in thetable 5-3. When switches S1 in position B and S2 in position D. Then, we record themagnitude and the direction of the current as indicated by each ammeter in appropriatetable and repeat the step above with Vn replace by Vn’’. Don’t forget to refer to the third
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configuration in figure 5-4, Last but not least, we simulate the ammeter and voltmeter for each switching combination used during the test in items 2, 4 and 5 by using themeasured values of the voltages and the resistance. Then, fill in the table 5-3. Thecurrent and the voltages are displays on the schematics.. From analysis, we examine theoutput acquire the branch currents’ direction. The theoretical value is record using P-
spices.
CIRCUIT DIAGRAMS
Figure 5-3
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Figure 5-4 (ii)
Figure 5-4 (ii)
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Figure 5-4 (iii)
RESULT
Measured values.
V1(V)
V2(V)
R 1(Ω)
R 2(kΩ)
R 3(kΩ)
12 6 465 5.51 3.58
Table 5.2
SwitchPositions
Ammeter readingsMeasured Simulated
S1 S2 A1(mA)
A2(mA)
A3(mA)
Nodevoltage
A1(mA)
A2(mA)
A3(mA)
Nodevoltage
A D 5.85 -1.76 -4.26 Vn =-9.28 5.881 1.649 4.232 Vn =-9.236
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B D 1.36 0.158 -1.57 Vn’ =-676 0.1558 1.372 1.488 Vn’ =-644.99
A C -4.46 -1.71 -2.74 Vn” =-9.92 4.509 1.764 2.745 Vn” =-9.88
Table 5.3
Percentage error:
A1: A2:
|5.86-5.881| x 100% =0.35% |1.67-1.649| x 100% = 0.7%5.881 1.649
|0.155-0.1558| x 100% = 0.51% |1.36-1.372| x 100% = 0.87%0.1558 1.372
|4.51-4.509| x 100% = 0.1% |1.783-1.768| x 100% = 1.69%4.509 1.768
A3:
|4.2-4.232| x 100% = 0.76%4.232
|1.48-1.488| x 100% = 0.53%1.488
|2.74-2.745| x 100% = 0.18%2.745
DISCUSSION AND CONCLUSION.Regarding to the experiment, the superposition principle state that the voltage
across ( or the current through ) an element in a linear circuit is algebraic sum of thevoltages across ( or current through) that element due to each independent sourceacting alone. From the experiment conducted, we can conclude that we have verifiedthe aims for this experiment. Also, it is based on linearity. For this reason, the power
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absorbed by a resistor depends on the square of the voltage/current. Thus,mathematically
A1 + A2 + A3 = 0
In this case, considered two limiting factors in circuit arrangement which wouldtend to make the superposition theorem unsatisfactory as a method of analysis. Thesuperposition theorem can’t be applied to a circuit:
• If the circuit has more than one independent circuit at a time. Because, wemust consider one independent source at a time while all other independent source are turned off. This implies that we replace everyvoltage source by 0 V (or short circuit), and every current source by 0 A(or an open circuit)
• If the dependent source are not connected to the circuit. Dependentsources should be left intact because they are controlled by circuit
variables.